Tan half angle formula

\( \newcommand\) • • • • • • Solve sine, cosine, and tangent of angles multiplied or divided by 2. Trig Riddle: I am an angle x such that \(0\leq x<2\pi \). I satisfy the equation \(\sin 2x−\sin x=0\). What angle am I? Solve Trigonometric Equations We can use the half and double angle formulas to solve trigonometric equations. Let's solve the following trigonometric equations. • Solve \(\tan 2x+\tan x=0\) when \(0\leq x<2\pi \). Change \(\tan 2x\) and simplify. \(\begin Review Solve the following equations for \(0\leq x<2\pi \). • \(\cos x−\cos \dfrac=1\) • \(\cos 2x−1=\sin ^2x\) • \(\cos 2x=\cos x\) • \(\sin 2x−\cos 2x=1\) • \(\sin ^2x−2=\cos 2x\) • \(\cot x+\tan x=2\csc 2x\)

tan ⁡ 1 2 ( η ± θ ) = tan ⁡ 1 2 η ± tan ⁡ 1 2 θ 1 ∓ tan ⁡ 1 2 η tan ⁡ 1 2 θ = sin ⁡ η ± sin ⁡ θ cos ⁡ η + cos ⁡ θ = − cos ⁡ η − cos ⁡ θ sin ⁡ η ∓ sin ⁡ θ , tan ⁡ 1 2 θ = sin ⁡ θ 1 + cos ⁡ θ = tan ⁡ θ sec ⁡ θ + 1 = 1 csc ⁡ θ + cot ⁡ θ , ( η = 0 ) tan ⁡ 1 2 θ = 1 − cos ⁡ θ sin ⁡ θ = sec ⁡ θ − 1 tan ⁡ θ = csc ⁡ θ − cot ⁡ θ , ( η = 0 ) tan ⁡ 1 2 ( θ ± 1 2 π ) = 1 ± sin ⁡ θ cos ⁡ θ = sec ⁡ θ ± tan ⁡ θ = csc ⁡ θ ± 1 cot ⁡ θ , ( η = 1 2 π ) tan ⁡ 1 2 ( θ ± 1 2 π ) = cos ⁡ θ 1 ∓ sin ⁡ θ = 1 sec ⁡ θ ∓ tan ⁡ θ = cot ⁡ θ csc ⁡ θ ∓ 1 , ( η = 1 2 π ) 1 − tan ⁡ 1 2 θ 1 + tan ⁡ 1 2 θ = ± 1 − sin ⁡ θ 1 + sin ⁡ θ tan ⁡ 1 2 θ = ± 1 − cos ⁡ θ 1 + cos ⁡ θ From these one can derive identities expressing the sine, cosine, and tangent as functions of tangents of half-angles: sin ⁡ α = 2 tan ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α cos ⁡ α = 1 − tan 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α tan ⁡ α = 2 tan ⁡ 1 2 α 1 − tan 2 ⁡ 1 2 α gives sin ⁡ α = 2 sin ⁡ 1 2 α cos ⁡ 1 2 α = 2 sin ⁡ 1 2 α cos ⁡ 1 2 α / cos 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α = 2 tan ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α , and cos ⁡ α = cos 2 ⁡ 1 2 α − sin 2 ⁡ 1 2 α = ( cos 2 ⁡ 1 2 α − sin 2 ⁡ 1 2 α ) / cos 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α = 1 − tan 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α , and Taking the quotient of the formulae for sine and cosine yields tan ⁡ α = 2 tan ⁡ 1 2 α 1 − tan 2 ⁡ 1 2 α . rearranging, and taking the square roots yields | tan ⁡ α | = 1 − cos ⁡ 2 α 1 + cos ⁡ 2 α = 1 − cos ⁡ 2 α 1 + cos ⁡ 2 α 1 + cos ⁡ 2 α = 1 − cos 2 ⁡ 2 α 1 + cos ⁡ 2 α = | sin ⁡ 2 α | 1 + cos ⁡ 2 ...

Half angle formulae for #(tan(theta))/2# #color(white)("XXX")=(sin(theta))/(1+cos(theta))color(white)("xxxxxxxx")#[1] #color(white)("XXX")=(cos(theta))/(1-sin(theta))color(white)("xxxxxxxx")#[2] #color(white)("XXX")=+-sqrt(1-cos(theta))/(1+cos(theta))color(white)("xxxx")#[3] Since #sin(30^circ)=1/2color(white)("xx")#and #color(white)("xx")cos(30^circ)=sqrt(3)/2# We can use [2] (for example) to get #tan(15^circ)=tan((30^circ)/2)=((sqrt(3)/2))/((1-1/2))=sqrt(3)#

. Proof Consider a semi-circle with “center” and diameter and radius equal to 1 unit as shown below. If we let , then by the . Draw perpendicular to as shown in the second figure. We can compute for the sine and cosine of which equal to the lengths of and , respectively. In effect, and . Draw . Notice that and are similar triangles, so their corresponding angles are congruent. So, . Now, we compute for the tangent of . In triangle , . In triangle , . Therefore, . And we are done. *** The last figure is the proof without words of R. J. Walker.

Half-Angle Formulae For finding the values of angles apart from the well-known values of 0°, 30°, 45°, 60°, 90°, and 180°. Half angles are derived from double angle formulas and are listed below for sin, cos, and tan: • sin (x/2) = ± [(1 – cos x)/ 2] 1/2 • cos (x/2) = ± [(1 + cos x)/ 2] 1/2 • tan (x/ 2) = (1 – cos x)/ sin x • Half Angle Formula of Tan, tan A/2 = ±√[1 – cos A] / [1 + cos A] tan A/2 = sin A / (1 + cos A) tan A/2 = (1 – cos A) / sin A Half Angle Formulas Derivation Using Double Angle Formulas Half-Angle formulas are derived using double-angle formulas. Before learning about half-angle formulas we must learn about Double-angle in Trigonometry, most commonly used double-angle formulas in trigonometry are: • sin 2x = 2 sin x cos x • cos 2x = cos 2 x – sin 2 x = 1 – 2 sin 2x = 2 cos 2x – 1 • tan 2x = 2 tan x / (1 – tan 2x) Now replacing x with x/2 on both sides in the above formulas we get • sin x = 2 sin(x/2) cos(x/2) • cos x = cos 2 (x/2) – sin 2 (x/2) = 1 – 2 sin 2 (x/2) = 2 cos 2(x/2) – 1 • tan A = 2 tan (x/2) / [1 – tan 2(x/2)] Half-Angle Formula for Cos Derivation We use cos2x = 2cos 2x – 1 for finding the Half-Angle Formula for Cos Put x = 2y in the above formula cos (2)(y/2) = 2cos 2(y/2) – 1 cos y = 2cos 2(y/2) – 1 1 + cos y = 2cos 2(y/2) 2cos 2(y/2) = 1 + cosy cos 2(y/2) = (1+ cosy)/2 cos(y/2) = ± √ Half-Angle Formula for Tan Derivation We know that tan x = sin x / cos x such that, tan(x/2) = sin(x/2) / cos(x/2) Putting the values of half angle for sin ...

. Proof Consider a semi-circle with “center” and diameter and radius equal to 1 unit as shown below. If we let , then by the . Draw perpendicular to as shown in the second figure. We can compute for the sine and cosine of which equal to the lengths of and , respectively. In effect, and . Draw . Notice that and are similar triangles, so their corresponding angles are congruent. So, . Now, we compute for the tangent of . In triangle , . In triangle , . Therefore, . And we are done. *** The last figure is the proof without words of R. J. Walker.

tan ⁡ 1 2 ( η ± θ ) = tan ⁡ 1 2 η ± tan ⁡ 1 2 θ 1 ∓ tan ⁡ 1 2 η tan ⁡ 1 2 θ = sin ⁡ η ± sin ⁡ θ cos ⁡ η + cos ⁡ θ = − cos ⁡ η − cos ⁡ θ sin ⁡ η ∓ sin ⁡ θ , tan ⁡ 1 2 θ = sin ⁡ θ 1 + cos ⁡ θ = tan ⁡ θ sec ⁡ θ + 1 = 1 csc ⁡ θ + cot ⁡ θ , ( η = 0 ) tan ⁡ 1 2 θ = 1 − cos ⁡ θ sin ⁡ θ = sec ⁡ θ − 1 tan ⁡ θ = csc ⁡ θ − cot ⁡ θ , ( η = 0 ) tan ⁡ 1 2 ( θ ± 1 2 π ) = 1 ± sin ⁡ θ cos ⁡ θ = sec ⁡ θ ± tan ⁡ θ = csc ⁡ θ ± 1 cot ⁡ θ , ( η = 1 2 π ) tan ⁡ 1 2 ( θ ± 1 2 π ) = cos ⁡ θ 1 ∓ sin ⁡ θ = 1 sec ⁡ θ ∓ tan ⁡ θ = cot ⁡ θ csc ⁡ θ ∓ 1 , ( η = 1 2 π ) 1 − tan ⁡ 1 2 θ 1 + tan ⁡ 1 2 θ = ± 1 − sin ⁡ θ 1 + sin ⁡ θ tan ⁡ 1 2 θ = ± 1 − cos ⁡ θ 1 + cos ⁡ θ From these one can derive identities expressing the sine, cosine, and tangent as functions of tangents of half-angles: sin ⁡ α = 2 tan ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α cos ⁡ α = 1 − tan 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α tan ⁡ α = 2 tan ⁡ 1 2 α 1 − tan 2 ⁡ 1 2 α gives sin ⁡ α = 2 sin ⁡ 1 2 α cos ⁡ 1 2 α = 2 sin ⁡ 1 2 α cos ⁡ 1 2 α / cos 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α = 2 tan ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α , and cos ⁡ α = cos 2 ⁡ 1 2 α − sin 2 ⁡ 1 2 α = ( cos 2 ⁡ 1 2 α − sin 2 ⁡ 1 2 α ) / cos 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α = 1 − tan 2 ⁡ 1 2 α 1 + tan 2 ⁡ 1 2 α , and Taking the quotient of the formulae for sine and cosine yields tan ⁡ α = 2 tan ⁡ 1 2 α 1 − tan 2 ⁡ 1 2 α . rearranging, and taking the square roots yields | tan ⁡ α | = 1 − cos ⁡ 2 α 1 + cos ⁡ 2 α = 1 − cos ⁡ 2 α 1 + cos ⁡ 2 α 1 + cos ⁡ 2 α = 1 − cos 2 ⁡ 2 α 1 + cos ⁡ 2 α = | sin ⁡ 2 α | 1 + cos ⁡ 2 ...

\( \newcommand\) • • • • • • Solve sine, cosine, and tangent of angles multiplied or divided by 2. Trig Riddle: I am an angle x such that \(0\leq x<2\pi \). I satisfy the equation \(\sin 2x−\sin x=0\). What angle am I? Solve Trigonometric Equations We can use the half and double angle formulas to solve trigonometric equations. Let's solve the following trigonometric equations. • Solve \(\tan 2x+\tan x=0\) when \(0\leq x<2\pi \). Change \(\tan 2x\) and simplify. \(\begin Review Solve the following equations for \(0\leq x<2\pi \). • \(\cos x−\cos \dfrac=1\) • \(\cos 2x−1=\sin ^2x\) • \(\cos 2x=\cos x\) • \(\sin 2x−\cos 2x=1\) • \(\sin ^2x−2=\cos 2x\) • \(\cot x+\tan x=2\csc 2x\)

Half-Angle Formulae For finding the values of angles apart from the well-known values of 0°, 30°, 45°, 60°, 90°, and 180°. Half angles are derived from double angle formulas and are listed below for sin, cos, and tan: • sin (x/2) = ± [(1 – cos x)/ 2] 1/2 • cos (x/2) = ± [(1 + cos x)/ 2] 1/2 • tan (x/ 2) = (1 – cos x)/ sin x • Half Angle Formula of Tan, tan A/2 = ±√[1 – cos A] / [1 + cos A] tan A/2 = sin A / (1 + cos A) tan A/2 = (1 – cos A) / sin A Half Angle Formulas Derivation Using Double Angle Formulas Half-Angle formulas are derived using double-angle formulas. Before learning about half-angle formulas we must learn about Double-angle in Trigonometry, most commonly used double-angle formulas in trigonometry are: • sin 2x = 2 sin x cos x • cos 2x = cos 2 x – sin 2 x = 1 – 2 sin 2x = 2 cos 2x – 1 • tan 2x = 2 tan x / (1 – tan 2x) Now replacing x with x/2 on both sides in the above formulas we get • sin x = 2 sin(x/2) cos(x/2) • cos x = cos 2 (x/2) – sin 2 (x/2) = 1 – 2 sin 2 (x/2) = 2 cos 2(x/2) – 1 • tan A = 2 tan (x/2) / [1 – tan 2(x/2)] Half-Angle Formula for Cos Derivation We use cos2x = 2cos 2x – 1 for finding the Half-Angle Formula for Cos Put x = 2y in the above formula cos (2)(y/2) = 2cos 2(y/2) – 1 cos y = 2cos 2(y/2) – 1 1 + cos y = 2cos 2(y/2) 2cos 2(y/2) = 1 + cosy cos 2(y/2) = (1+ cosy)/2 cos(y/2) = ± √ Half-Angle Formula for Tan Derivation We know that tan x = sin x / cos x such that, tan(x/2) = sin(x/2) / cos(x/2) Putting the values of half angle for sin ...

Half angle formulae for #(tan(theta))/2# #color(white)("XXX")=(sin(theta))/(1+cos(theta))color(white)("xxxxxxxx")#[1] #color(white)("XXX")=(cos(theta))/(1-sin(theta))color(white)("xxxxxxxx")#[2] #color(white)("XXX")=+-sqrt(1-cos(theta))/(1+cos(theta))color(white)("xxxx")#[3] Since #sin(30^circ)=1/2color(white)("xx")#and #color(white)("xx")cos(30^circ)=sqrt(3)/2# We can use [2] (for example) to get #tan(15^circ)=tan((30^circ)/2)=((sqrt(3)/2))/((1-1/2))=sqrt(3)#